= Solution
For a downward <step potential>, $V_0=-|V_0|$, the region on the right is classically allowed at every positive incident <energy>. The same matching calculation applies with
$$
k_1=\frac{\sqrt{2mE}}\hbar,\qquad
k_2=\frac{\sqrt{2m(E+|V_0|)}}\hbar.
$$
As $E\to0^+$, $k_1\to0$ and $k_2\to k_*=\sqrt{2m|V_0|}/\hbar>0$. Hence
$$
\frac BA=\frac{k_1-k_2}{k_1+k_2}\to-1,\qquad
\frac CA=\frac{2k_1}{k_1+k_2}\sim\frac{2k_1}{k_*}\to0,
$$
and the flux coefficients satisfy
$$
\boxed{R=\left(\frac{k_1-k_2}{k_1+k_2}\right)^2\longrightarrow1,\qquad
T=\frac{4k_1k_2}{(k_1+k_2)^2}\sim\frac{4k_1}{k_*}\longrightarrow0.}
$$
This <quantum reflection at a downward step> occurs even though a classical particle would simply accelerate into the lower-potential region. The abrupt change in <wavenumber> causes strong reflected-wave interference.
Both this limit and part (d) give total reflection, but their <probability amplitudes> and spatial densities differ. At the upward-step threshold, $B/A\to+1$ and $C/A\to2$: an antinode meets the origin and the right-hand density stays finite while its propagation <velocity> vanishes. Here $B/A\to-1$ and $C/A\to0$: the incident and reflected waves nearly cancel at the origin, and the transmitted density vanishes while the right-hand propagation velocity remains finite. In both descriptions the incident <probability amplitude> is held fixed for comparison.
The coefficients above are limiting flux ratios for $E>0$. At exactly $E=0$ the incident <probability current> vanishes, so directly dividing zero-energy fluxes would not define a scattering probability.
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